我使用这个函数

function getDatesRange(startDate, stopDate) {
    const ONE_DAY = 24*3600*1000;
    var days= [];
    var currentDate = new Date(startDate);
    while (currentDate <= stopDate) {
        days.push(new Date (currentDate));
        currentDate = currentDate - 1 + 1 + ONE_DAY;
    }
    return days;
}

Date.prototype.addDays = function(days) {
    var date = new Date(this.valueOf());
    date.setDate(date.getDate() + days);
    return date;
}

function getDates(startDate, stopDate) {
    var dateArray = new Array();
    var currentDate = startDate;
    while (currentDate <= stopDate) {
        dateArray.push(new Date (currentDate));
        currentDate = currentDate.addDays(1);
    }
    return dateArray;
}

这里是一个功能演示http://jsfiddle.net/jfhartsock/cM3ZU/

我最近在用moment.js工作，下面做了一个戏法。

function getDateRange(startDate, endDate, dateFormat) {
        var dates = [],
            end = moment(endDate),
            diff = endDate.diff(startDate, 'days');

        if(!startDate.isValid() || !endDate.isValid() || diff <= 0) {
            return;
        }

        for(var i = 0; i < diff; i++) {
            dates.push(end.subtract(1,'d').format(dateFormat));
        }

        return dates;
    };
    console.log(getDateRange(startDate, endDate, dateFormat));

结果将是:

["09/03/2015", "10/03/2015", "11/03/2015", "12/03/2015", "13/03/2015", "14/03/2015", "15/03/2015", "16/03/2015", "17/03/2015", "18/03/2015"]

不是最短的，而是简单的，不可变的，没有依赖关系

function datesArray(start, end) {
    let result = [], current = new Date(start);
    while (current <= end)
        result.push(current) && (current = new Date(current)) && current.setDate(current.getDate() + 1);
    return result;
}

使用

函数datesArray(start, end) { let result = []， current = new Date(start); While (current <= end) result.push(current) && (current = new Date(current)) && current. setdate (current. getdate () + 1); 返回结果; } / /使用 const test = datesArray( 新的日期(“2020-02-26”), 新日期(“2020-03-05”) ）; 对于(设I = 0;I < test.length;I ++) { console.log ( 测试[我].toISOString () .slice (0, 10) ）; }

var listDate = [];
var startDate ='2017-02-01';
var endDate = '2017-02-10';
var dateMove = new Date(startDate);
var strDate = startDate;

while (strDate < endDate){
  var strDate = dateMove.toISOString().slice(0,10);
  listDate.push(strDate);
  dateMove.setDate(dateMove.getDate()+1);
};
console.log(listDate);

//["2017-02-01", "2017-02-02", "2017-02-03", "2017-02-04", "2017-02-05", "2017-02-06", "2017-02-07", "2017-02-08", "2017-02-09", "2017-02-10"]

我看了上面所有的。最后我自己开始写作。你不需要花时间。原生的for循环就足够了，而且最有意义，因为for循环的存在是为了计算一个范围内的值。

一个衬套:

var getDaysArray = function(s,e) {for(var a=[],d=new Date(s);d<=new Date(e);d.setDate(d.getDate()+1)){ a.push(new Date(d));}return a;};

长版本

var getDaysArray = function(start, end) {
    for(var arr=[],dt=new Date(start); dt<=new Date(end); dt.setDate(dt.getDate()+1)){
        arr.push(new Date(dt));
    }
    return arr;
};

列出中间的日期:

var daylist = getDaysArray(new Date("2018-05-01"),new Date("2018-07-01"));
daylist.map((v)=>v.toISOString().slice(0,10)).join("")
/*
Output: 
    "2018-05-01
    2018-05-02
    2018-05-03
    ...
    2018-06-30
    2018-07-01"
*/

从过去日期到现在的日子:

var daylist = getDaysArray(new Date("2018-05-01"),new Date());
daylist.map((v)=>v.toISOString().slice(0,10)).join("")